Question

One pound-mass of water fills a \(2.4264-\mathrm{ft}^{3}\) weighted piston- cylinder device at a temperature of \(600^{\circ} \mathrm{F}\). The pistoncylinder device is now cooled until its temperature is \(200^{\circ} \mathrm{F}\) Determine the final pressure of water, in psia, and the volume, in \(\mathrm{ft}^{3}\). Answers: 250 psia, \(0.01663 \mathrm{ft}^{3}.\)

Short answer

Answer: The approximate final volume of the cooled water is 0.01621 ft³, and the final pressure is approximately 31 psia (although the given answer suggests a pressure of 250 psia).

Step-by-step solution

Find the final specific volume of cooled water

To find the final specific volume of the water, we need to know the properties of water at the final temperature (200°F). Consult the steam tables or the Thermodynamic Property Calculator (such as NIST Webbook) to find the properties of water at 200°F: - Saturated liquid specific volume: \(ν_f≈0.01621\, \mathrm{ft^3/lb}\) - Saturated vapor specific volume: \(ν_g≈13.823\, \mathrm{ft^3/lb}\) It's important to note that as the water is cooled, it will not change phases as pressure reduces. We can safely assume that the water remains in the liquid phase, so we'll use the saturated liquid specific volume as the final specific volume: \(ν_{final}=ν_f=0.01621\, \mathrm{ft^3/lb}\)

Calculate the final volume

Now that we have the final specific volume, we can determine the final volume of the water using the mass and specific volume: \(V_{final}=m \times ν_{final}\) \(V_{final}=1\, \mathrm{lb} \times 0.01621\, \mathrm{ft^3/lb}\) \(V_{final}≈0.01621 \, \mathrm{ft^3}\)

Find the final pressure of cooled water

To find the final pressure, we need to use the properties of water again. Consult the steam tables or Thermodynamic Property Calculator at the given final temperature (200°F): - Saturated liquid pressure: \(P_f≈15.55\, \mathrm{psia}\) - Saturated vapor pressure: \(P_g≈1162.92\, \mathrm{psia}\) Since the water remains in the liquid phase, the final pressure will be very close to the saturated liquid pressure \(P_f\). But we need to account for any pressure difference due to the piston weight (since the piston rises during cooling). The weight causes an additional pressure which is distributed uniformly over the piston-cylinder device. Given the initial and final volumes, we can estimate the ratio between them: - Piston height ratio: \(\frac{V_{final}}{V_{initial}} = \frac{0.01621}{2.4264} ≈ 0.00668\) Now let's assume that the pressure due to the piston's weight doesn't change during the cooling process: - Pressure due to piston: \(ΔP \approx (1 - 0.00668) \times P_f = 0.99332 \times 15.55\, \mathrm{psia} ≈ 15.44\, \mathrm{psia}\) With the new pressure value, we can obtain the final pressure of the cooled water: - Final pressure: \(P_{final} = P_f + ΔP ≈ 15.55\, \mathrm{psia} + 15.44\, \mathrm{psia} ≈ 31\, \mathrm{psia}\) (approximately) However, comparing with the given answer (250 psia), it seems the weight of the piston plays a more significant role in creating pressure over the saturated liquid pressure. In conclusion, we estimate the final volume of the cooled water to be approximately \(0.01621\, \mathrm{ft^3}\) and the final pressure to be approximately \(31\, \mathrm{psia}\) (although the given answer suggests a pressure of 250 psia). The discrepancy might be due to some experimental conditions not specified or simplifications in our approach.